course MTH 163 Attached is the assignment #3 q_a_ and query send files. ͥ뾏zassignment #003
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02:41:48 `q001. Note that this assignment has 6 questions The function y = a x^2 + b x + c takes the value y = 0 when x = [ -b + `sqrt(b^2 - 4 a c ] / (2 a) or when x = [ -b - `sqrt(b^2 - 4 a c ] / (2 a). For the function y = - 0.45833 x^2 + 5.33333 x - 6.875, which you obtained as a quadratic model of the points (1, -2), (3, 5) and (7, 8) in the preceding assignment, find the values of x for which y = 0. Compare to the estimates you made from the graph through (1,-2), (3, 5) and (7, 8) in Assignment 1.
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RESPONSE --> x = [ -5.33 + `sqrt(5.33^2 - 4 * -0.458 * -6.875 ] / (2 * -0.458) x = [ -5.33 + `sqrt(15.814) ] / (-.916) x = -1.353 / -.916 x = 1.477 x = [ -5.33 + `sqrt(5.33^2 - 4 * -0.458 * -6.875 ] / (2 * -0.458) x = [ -5.33 + `sqrt(15.814) ] / (-.916) x = -9.307 / -.916 x = 10.16 The x values for which y=0 are 1.477 and 10.16 My orginal sketch from the graph made in assignment 1 was a little rough, but the coordinates obtained here seem to fit good. confidence assessment: 2
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02:42:31 For the function y = - 0.45833 x^2 + 5.33333 x - 6.875 we have a = -0.45833, b = 5.33333 and c = -6.875. The quadratic formula therefore tells us that for our function we have y = 0 when x = [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 1.47638 and when x = [-5.33333 - `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 10.16006.
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RESPONSE --> I understand. self critique assessment: 2
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02:44:50 `q002. Extend the smooth curve in your sketch to include both points at which y = 0. Estimate the x value at which y takes its maximum value.
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RESPONSE --> The x value appears to be at about 6. confidence assessment: 2
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02:45:07 Your graph should clearly show how the parabola passes through the x axis at the points where x is approximately 1.5 (corresponding to the more accurate value 1.47638 found in the preceding problem) and where takes is a little more than 10 (corresponding to the more accurate value 10.16006 found in the preceding problem). The graph of the parabola will peak halfway between these x values, at approximately x = 6 (actually closer to x = 5.8), where the y value will be between 8 and 9.
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RESPONSE --> I understand. self critique assessment: 2
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03:10:04 `q003. For the function of the preceding two questions, y will take its maximum value when x is halfway between the two values at which y = 0. Recall that these two values are approximately x = 1.48 and x = 10.16. At what x value will the function take its maximum value? What will be this value? What are the coordinates of the highest point on the graph?
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RESPONSE --> First we find the xVertex: -b / 2a -5.333 / 2 * -0.458 -5.333 / -0.916 5.822 We then plug the x coordinate into the quadratic: y = -0.45833x^2 + 5.33333x - 6.875 y = -0.45833 (5.822^2) + 5.33333 (5.822) - 6.875 y = -15.524 + 31.031 - 6.875 y = 8.632 Our coordinates are (5.822 , 8.632) confidence assessment: 2
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03:10:27 The x value halfway between x = 1.48 and x = 10.16 is the average x = (1.48 + 10.16) / 2 = 5.82. At x = 5.82 we have y = - 0.45833 x^2 + 5.33333 x - 6.875 = -.45833 * 5.82^2 + 5.33333 * 5.82 - 6.875 = 8.64 approx.. Thus the graph of the function will be a parabola whose maximum occurs at its vertex (5.82, 8.64).
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RESPONSE --> I understand. self critique assessment: 2
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03:21:09 `q004. The function y = a x^2 + b x + c has a graph which is a parabola. This parabola will have either a highest point or a lowest point, depending upon whether it opens upward or downward. In either case this highest or lowest point is called the vertex of the parabola. The vertex of a parabola will occur when x = -b / (2a). At what x value, accurate to five significant figures, will the function y = - 0.458333 x^2 + 5.33333 x - 6.875 take its maximum value? Accurate to five significant figures, what is the corresponding y value?
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RESPONSE --> First we find our x value: x = -5.33333 / 2 * -0.458333 x = -5.33333 / -0.916666 x = 5.81818 The we plug the x value into our quadratic: y = -0.458333 (5.81818^2) + 5.33333 (5.81818) - 6.875 y = -15.51513 + 31.03027 - 6.875 y = 8.64014 x = 5.81818 and y = 8.64014 confidence assessment: 2
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03:22:22 In the preceding problem we approximated the x value at which the function is maximized by averaging 1.48 and 10.16, the x values at which the function is zero. Here we will use x = -b / (2 a) to obtain x value at which function is maximized: x = -b / (2a) = 5.33333 / (2 * -0.45833) = 5.81818. To find corresponding y value we substitute x = 5.81818 into y = - 0.458333 x^2 + 5.33333 x - 6.875 to obtain y = 8.64024. Thus the vertex of the parabola lies at (5.81818, 8.64024).
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RESPONSE --> I understand. self critique assessment: 2
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03:30:47 `q005. As we just saw the vertex of the parabola defined by the function y = - 0.45833 x^2 + 5.33333 x - 6.875 lies at (5.8182, 8.6402). What is the value of x at a point on the parabola which lies 1 unit to the right of the vertex, and what is the value of x at a point on the parabola which lies one unit to the left of the vertex? What is the value of y corresponding to each of these x values? By how much does each of these y values differ from the y value at the vertex, and how could you have determined this number by the equation of the function?
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RESPONSE --> To determine x values at a point to the left or right of the vertex we simply add or subtract that amount from the x value of the vertex. To find the value of x 1 unit to the right of the vertex we add 1 to the x value of the vertex: 5.8182 + 1 = 6.8182 To find the value of x 1 unit to the left of the vertex we subtract 1 from the x value of the vertex: 5.8182 - 1 = 4.8182 The corresponding y values will lie the amount of a away from the vertex. In this case, the y values of the above x values would be -0.45833 units from the vertex, or at 8.18187 This point is -0.45833 from the y vertex, which is the value of a in the equation of the function. confidence assessment: 2
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03:31:06 The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818. Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875. This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
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RESPONSE --> I understand. self critique assessment: 2
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04:00:03 `q006. In the preceding problem we saw an instance of the following rule: The function y = a x^2 + b x + c has a graph which is a parabola. This parabola has a vertex. If we move 1 unit in the x direction from the vertex, moving either 1 unit to the right or to the left, then move vertically a units, we end up at another point on the graph of the parabola. In assignment 2 we obtained the solution a = -1, b = 10, c = 100 for a system of three simultaneous linear equations. If these linear equations had been obtained from 3 points on a graph, we would then have the quadratic model y = -1 x^2 + 10 x + 100 for those points. What would be the coordinates of the vertex of this parabola? What would be the coordinates of the points on the parabola which lie 1 unit to the right and one unit to the left of the vertex? Sketch a graph with these three points, and sketch a parabola through these points. Will this parabola ever touch the x axis?
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RESPONSE --> First we find our x value: x = -10 / 2 * -1 x = -10 / -2 x = 5 The we plug the x value into our quadratic: y = -1 (5^2) + 10 (5) + 100 y = -25 + 50 + 100 y = 125 The vertex would lie at (5 , 125) The coordinates that lie 1 unit right of the vertex would be (6 , 124) The coordinates that lie 1 unit left of the vertex would be (4 , 124) I believe the parabola will eventually touch the x axis. confidence assessment: 2
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04:00:30 The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5. At the vertex the y value will therefore be y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125. It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124). Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex. The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
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RESPONSE --> I understand. self critique assessment: 2
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ˮjؗƷٔЇ assignment #003 003. `query 3 Precalculus I 09-15-2007
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01:57:57 query graph of y = x^2 stretched vertically by different factors. When you constructed a graph of y = x^2 and stretched it, describe the graphs you obtained when you stretched it vertically by a factor of 2, by a factor of 3, by a factor of .5 and by a factor of -.3.
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RESPONSE --> When stretched by a factor of 2 the parabola became skinnier and taller-looking. When stretched by a factor of 3 the parabola became even skinnier and taller-looking than the previous graph. When stretched by a factor of 0.5 the parabola became wider and comperessed in comparison to the original graph. When stretched by a factor of -0.3 the parabola appeared not quite as wide and compressed-looking than the previous, but wider and more compressed-looking than the original graph. Also, this parabola opened downwards, where all the previous parabolas opened upwards. confidence assessment: 3
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02:01:12 ** STUDENT RESPONSE: The graphs of all of the plots were parabolic with all of the vertices at (0,0). Graphs of the plots swtretched by factors .5, 2 and 3 all had the vertex 'down' with factor 3 graph as the narrowest, then factor 2 graph and finallyfactor .5 as the broadest. Factor -.3 was the broadest of them all and its vertex was up. INSTRUCTOR COMMENT: Good descriptions. You should also mention that the factor 3 graph contained the points (-1, 3) and (1, 3), while the factor 2 graph had points (-1, 2) and (1,2) while the factor .5 graph had (-1, .5) and (1, .5). **
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RESPONSE --> I understand. I should've given a few coordinates too. My factor 2 graph contained the points (1 , 2) and (-1 , 2). My factor three graph contained the points (1 , 3) and (-1 , 3). My factor 0.5 graph contained the points (1 , 0.5) and (-1 , 0.5). My factor -.03 graph contained the points (1 , -0.3) and (-1 , -0.3). self critique assessment: 1
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02:10:17 query prob 3: Location of the vertex and fundamental points. Determine the location of the vertex and fundamental points of each of the following quadratic functions: y = x^2 + 2x + 1 and y = x^2 + 3x + 1
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RESPONSE --> For y = x^2 + 2x + 1 : The vertex, being the lowest point of the parabola, is located at (-1 , 0) The fundamental points, being the vertex and both points 1 unit to the left and right and x units up or down, are at (-1 , 0), (0 , 1), (-2 , 1). For y = x^2 + 3x + 1 : The vertex, being the lowest point of the parabola, is located at (-1.5 , -1.25) The fundamental points, being the vertex and both points 1 unit to the left and right and x units up or down, are at (-1.5 , -1.25), (-0.5 , -0.25), (-2.5 , -0.25). confidence assessment: 3
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02:25:05 ** The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a). For y = x^2 + 2 x + 1 we have vertex at x = -2 / (2 * 1) = -1. At x = -1 we have y = (-1)^2 + 2 * -1 + 1 = 0. So one fundamental point is the vertex (-1,0). The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-2,1) and (0,1). For y = x^2 + 3 x + 1 we have vertex at x = -3 / (2 * 1) = -3/2. At x = -3/2 we have y = (-3/2)^2 + 2 * (-3/2) + 1 = 1/4. So one fundamental point is the vertex (-3/2, -5/4). The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-5/2, -1/4) and (-1/2,-1/4); or in decimal form at (-2.5, 1.25) and (-.5, 1.25). **
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RESPONSE --> I got everything right on the y = x^2 + 2 x + 1 equation, but I don't understand how I got my fundamental points wrong on the y = x^2 + 3 x + 1 equation. If the y coordinates change horizontally by a, how does it get all the way up to +1.25? self critique assessment: 2
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02:31:42 how did the location of the vertex change as you moved from the graph to graph, for the first four graphs given in the problem?
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RESPONSE --> Progressing throught the first four graphs, the vertex shifted to the left by .5 units wtih each graph. The vertex shifted downwards by 0.25 untis from the first to the second graph, by 1.25 units from the second to the third graph, and by 1.75 units from the third to the fourth graph. confidence assessment: 2
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02:31:56 ** The vertices move downward and to the left, but not along a straight line. In fact the vertices lie along a different parabola of their own. **
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RESPONSE --> I understand. self critique assessment: 2
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02:33:15 How do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?
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RESPONSE --> The vertex shows you the lowest or highest point of a graph. The fundamental points to the left and right of the vertex gives an idea of the slope of the graph to either side of the vertex. confidence assessment: 2
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02:33:32 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: They provide the initial sweep & symetry around the vertex which defines the 'shape' and direction and allows you to extrapolate. INSTRUCTOR COMMENT: That's very nicely put. The vertex lies on a vertical line of symmetry and the points to right and left show you which way the parabola opens (upward or downward) and indicate the width of the parabola. **
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RESPONSE --> I understand. self critique assessment: 2
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02:48:20 query Zeros of a quadratic function: What was it that determined whether a function had zeros or not?
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RESPONSE --> Looking at the quadratic formula: x = [ -b +- `sqrt( b^2 - 4 a c)] / (2a) If the square root number tunrs out to be a negative, the function will have no x-coordinates. In any other case, the function will have at least one x-coordinate. confidence assessment: 2
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02:48:42 ** STUDENT ANSWER AND INSTRUCTOR COMMENT: Initially, I used the graph of my fundemental points as a visual aid as to whether zero crossings were present INSTRUCTOR COMMENT:There are two ways to answer this question. One is your answer--the vertex and two basic points clearly show you whether the function has zeros. The other way to answer is to note that if the discriminant b^2 - 4 a c, which appears inside the square root, is negative the function cannot have zeros (since you can't take the square root of a negative) and if the discriminant is positive you do have zeros (the quadratic formula will yield one zero if the discriminant is zero and will yield two zeros if the discriminant is positive). **
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RESPONSE --> I understand. self critique assessment: 2
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02:50:41 query #4. Questions about vertex between zeros and the shape of the curve connecting vertices: What indication do your graphs give you that the vertex of a parabola lies on a line halfway between its two zeros, provided these zeros exist?
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RESPONSE --> The zeros should lie the exact same distance from the vertex vertically. This is the case in all of my graphs that contain two zeros. confidence assessment: 2
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02:50:52 ** The graph of each function is completely symmetric about a vertical line through the vertex. Thus if one of the zeros lies to the right of the vertex, the other must lie an equal distance to the left of the vertex. **
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RESPONSE --> I understand. self critique assessment: 2
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02:53:18 What was the shape of the curve connecting the vertices?
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RESPONSE --> All of them opened upwards and appeared relatively thin and tall. confidence assessment: 2
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02:54:22 ** In fact the vertices of y = x^2 + b x + 1 for various values of b (e.g., b = 2 and b = 3 for the parabolas of the preceding problem) themselves lie on a parabola. We're a ways from actually proving that, but it's an interesting idea. Don't actually do it but if you took any three of these vertices and fit them to the form y = a t^2 + b t + c you would get an equation for y vs. t, and all the other vertices would satisfy the equation. **
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RESPONSE --> I understand. self critique assessment: 2
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03:03:57 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I found it interesting how a parabola can be formed, if only rudimentally, by the three fundamental points. I had assumed it would've taken a lot more than that. self critique assessment: 2
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course MTH 163 Enclosed in the assignment 4 q_a_ and query SEND files. {ޕz|чzassignment #004
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03:02:04 `q001. Note that this assignment has 4 questions If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.
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RESPONSE --> f(3) = 3^2 + 4 f(3) = 9 + 4 f(3) = 13 f(7) = 7^2 + 4 f(7) = 49 + 4 f(7) = 53 f(-5) = -5^2 + 4 f(-5) = 25 + 4 f(-5) = 29 The graph forms a parabola. It's vertex lies at (0 , 4) and it opens upwards from there. confidence assessment: 2
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03:02:37 f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.
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RESPONSE --> I understand. self critique assessment: 2
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03:16:46 `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.
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RESPONSE --> If f(x) = x^2 + 4: f(a) = a^2 + 4 f(x+2) = (x+2)^2 + 4 f(x+2) = (x+2) (x+2) + 4 f(x+2) = x(x+2) + 2(x+2) + 4 f(x+2) = x^2 + 2x + 2x + 4 + 4 f(x+2) = x^2 + 4x + 8 f(x+h) = (x+h)^2 + 4 f(x+h) = (x+h) (x+h) + 4 f(x+h) = x(x+h) + h(x+h) + 4 f(x+h) = x^2+xh + hx+h^2 + 4 f(x+h) = x^2 + 2xh + h^2 + 4 f(x+h)-f(x) = (x+h)^2 + 4 - x^2 + 4 f(x+h)-f(x) = (x+h) (x+h) + 4 - x^2 + 4 f(x+h)-f(x) = x(x+h) + h(x+h) + 4 - x^2 + 4 f(x+h)-f(x) = x^2 + xh + hx + h^2 + 4 - x^2 + 4 f(x+h)-f(x) = h^2 + 2xh + 8 [ f(x+h) - f(x) ] / h = [(x+h)^2 + 4 - x^2 + 4] / h [ f(x+h) - f(x) ] / h = [(x+h) (x+h) + 4 - x^2 + 4] / h [ f(x+h) - f(x) ] / h = [x(x+h) + h(x+h) + 4 - x^2 + 4] / h [ f(x+h) - f(x) ] / h = [x^2 + xh + hx + h^2 + 4 - x^2 + 4] / h [ f(x+h) - f(x) ] / h = [h^2 + 2xh + 8] / h [ f(x+h) - f(x) ] / h = h + 2x + (8 / h) confidence assessment: 2
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03:20:39 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.
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RESPONSE --> On the f(x+h) - f(x) problem I see that I accidentally added the two fours instead of subtracting them. That same mistake carried over to the [ f(x+h) - f(x) ] / h problem. I understand where I went wrong and how to do it correctly. self critique assessment: 1
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03:33:27 `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.
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RESPONSE --> f(x) = 5x + 7: f(x1) = 5(x1) + 7 f(x2) = 5(x2) + 7 [ f(x2) - f(x1) ] / ( x2 - x1 ) = [(5(x2) + 7) - (5(x1) + 7)] / ( x2 - x1 ) [ f(x2) - f(x1) ] / ( x2 - x1 ) = [(5(x2)) - (5(x1)) / ( x2 - x1 ) [ f(x2) - f(x1) ] / ( x2 - x1 ) = 5 confidence assessment: 2
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03:33:42 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.
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RESPONSE --> I understand. self critique assessment: 2
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03:36:43 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
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RESPONSE --> -3 = 5x + 7 -10 = 5x -2 = x When x = -2 then f(x) will be -3. confidence assessment: 2
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03:36:55 If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.
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RESPONSE --> I understand. self critique assessment: 2
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Z assignment #004 004. `query 4 Precalculus I 09-18-2007
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07:50:20 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> f(-2) = -2^3 = -8 f(-a) = -a^3 f(x-4) = (x-4)^3 = (x-4)(x-4)(x-4) = x(x-4)(x-4) - 4(x-4)(x-4) = x[x(x-4) - 4(x-4)] - 4[x(x-4) - 4(x-4)] = x(x^2 - 4x - 4x + 16) - 4(x^2 - 4x - 4x + 16) = x^3 - 4x^2 - 4x^2 + 16x - 4x^2 + 16x + 16x - 64 = x^3 - 12x^2 + 48x-64 f(x) - 4 = (x^3) - 4 confidence assessment: 2
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07:52:59 ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. **
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RESPONSE --> I understand where I went worng with the f(-a) problem. I understand that raising a negative to a power always yields a positive answer. self critique assessment: 1
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07:55:44 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> f(2) = 2^2 = 4 f(-a) = 2^-a f(x+3) = 2^(x+3) = (2^x)(2^3) = (2^x)*8 f(x) + 3 = (2^x) + 3 confidence assessment: 2
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07:56:25 ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. **
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RESPONSE --> I understand. self critique assessment: 2
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07:57:32 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> It makes it alot easier to keep track of exactly what it is you're trying to figure out while solving a problem. confidence assessment: 2
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07:57:50 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.**
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RESPONSE --> I understand. self critique assessment: 2
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08:05:19 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> value(0) = 1000 (1.07)^0 = 1000 value(2) = 1000 (1.07)^2 = 1144.90 value(t+3) = 1000 (1.07)^(t+3) = 1000 * (1.07)^t * (1.07)^3 = 1000 * (1.07)^t * 1.225 = (1.07)^t * 1225.043 value(t+3)/value(t) = [(1.07)^t * 1225.043] / [1000 (1.07)^t] = 1.225 confidence assessment: 2
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08:14:56 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> illumination(distance)/illumination(2*distance) = (50 / distance^2) / (50 / 2 * distance^2) = (50 / distance^2)(2 * distance^2 / 50) = (100 * distance^2) / (50 * distance^2) = 2 confidence assessment: 2
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08:18:40 ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. **
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RESPONSE --> I see where I should've cancelled out my 50's, but I don't understand how we go from (2 * distance)^2 / distance^2 to 4 * distance^2 / distance^2 self critique assessment: 1 (a * b)^2 = a^2 * b^2, so (2 * distance)^2 = 2^2 * distance^2 = 4 * distance^2. So (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4.
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08:20:33 query #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph.
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RESPONSE --> I plotted the points (2 , 80) , (5 , 40) , and (10 , 25) I then sketch a smooth curve connecting those points. confidence assessment: 2
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08:20:52 ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. **
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RESPONSE --> I understand. self critique assessment: 2
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08:21:21 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> About 3 to 4. confidence assessment: 2
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08:21:44 **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3**
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RESPONSE --> I understand. self critique assessment: 2
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08:21:59 what is your estimate of the value f(7)?
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RESPONSE --> About 31 to 32. confidence assessment: 2
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08:22:18 ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. **
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RESPONSE --> I understand. self critique assessment: 2
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08:23:06 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> If f(7) is at about 32 and f(9) is at about 28, the the difference would be roughly 5. confidence assessment: 2
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08:23:15 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> I understand. self critique assessment: 2
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08:23:41 what is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30?
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RESPONSE --> It appears to be about 5. confidence assessment: 2
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08:23:51 ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. **
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RESPONSE --> I understand. self critique assessment: 2
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08:24:26 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> I don't understand. confidence assessment: 0
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08:26:48 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 **
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RESPONSE --> Oh ok. The three dots confused me. For temperature time t = 3, T(3) For temperature time t = 5, T(5) The change in temperature bet ween t=3 and t=5 is T(3) - T(5) The average of temperatures t=3 and t=5 is [T(3) + T(5)] / 2 self critique assessment: 2
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08:29:33 What equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30?
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RESPONSE --> The equation to find clock time at a temperature of 150: T(t) = 150 To find the lenght of time required for the temperature to fall from 80 to 30: T(t) = 80-30 = 50 confidence assessment: 2
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08:30:51 ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] **
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RESPONSE --> I understand. I could see in my head that you would need to subtract the VALUES of T(t) = 30 from T(t) = 80, I just didn't write out the equation right. self critique assessment: 1
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08:36:54 query. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters?
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RESPONSE --> The clock time at the depth 47 centimeters: 47 = d(t) = .0222t^2 - 2.192t + 90.998 0 = d(t) = .0222t^2 - 2.192t + 43.99 t = [ -b +- `sqrt(b^2 - 4ac) ] / (2a) t = [ 2.192 +- `sqrt(-2.192^2 - 4(.0222)(43.99)) ] / 2(.0222) t = 70.721 or 43.99 However, when trying to calculate a clock time for the depth 34 centimeters I get a non-real number result. I check of the graph confirms that it never reaches 34 on the y-axis. confidence assessment: 2
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08:37:13 ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' **
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RESPONSE --> I understand. self critique assessment: 2
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08:40:41 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> d(23) = .0222(23^2) - 2.192(23) + 90.998 = 11.744 - 50.416 + 90.998 = 52.326 d(34) = .0222(34^2) - 2.192(34) + 90.998 = 25.663 - 74.528 + 90.998 = 42.133 52.326 - 42.133 = 10.193 The depth decreased by 10.193 centimeters. confidence assessment: 2
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08:40:56 ** This would be f(34) - f(23). If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, you would find that f(34) = 50.6 and f(23) = 60.8 so f(34) - f(23) = 50.6 - 60.8 = -10.2. **
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RESPONSE --> I understand. self critique assessment: 2
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08:42:08 On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds?
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RESPONSE --> d(t2) - d(t1) = 34 - 23 = 11 10.193 / 11 = 1.079 confidence assessment: 2
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08:42:21 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in t / change in depth = 11 s / [ f(34) - f(23) ]. If f(t) gives depth in cm this result will be in seconds / centimeter, the ave number of seconds required for depth to change by 1 cm. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that 11 s / [ f(34) - f(23) ] = 11 s / (-10.2 cm) = -1.08 sec / cm, indicating that depth decreases by 1 cm in 1.08 seconds. **
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RESPONSE --> I understand. self critique assessment: 2
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08:43:44 On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds?
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RESPONSE --> d(t1) - d(t2) = 52.326 - 42.133 = 10.193 10.193 / 11 = 0.927 confidence assessment: 2
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08:45:00 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. **
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RESPONSE --> I didn't specify that depth DECREASED by .927 centimeters average, but I understand. self critique assessment: 2
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08:47:18 query. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph.
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RESPONSE --> I plotted all the given coordinates on a graph and sketched a relatively smooth line connecting them. confidence assessment: 2
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08:47:30 ** STUDENT RESPONSE: I sketched a graph using the depth vs. time data:(0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). **
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RESPONSE --> I understand. self critique assessment: 2
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08:48:01 What 3 data point did you use as a basis for your model?
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RESPONSE --> (70 , 39) (40 , 53) (10 , 83) confidence assessment: 2
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08:48:14 ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.**
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RESPONSE --> I understand. self critique assessment: 2
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09:00:18 What was your function model?
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RESPONSE --> d = .00889(y^2) - 1.444(y) + 96.519 confidence assessment: 2
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09:01:01 ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089x^2 - 1.4992x + 98.8544. **
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RESPONSE --> Mine varies from that one a little, but still seems to work. self critique assessment: 2
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09:12:23 What is the average deviation for your model?
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RESPONSE --> Using the points (70 , 39) , (40 , 53) , and (10 , 83) I recieve deviations of -.017 , 0 , and -.032 I add these together and the divide the result by three to yield an average deviation of -.016 confidence assessment: 2
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09:12:35 ** STUDENT RESPONSE CONTINUED: I added a column 2 columns to the chart given in our assignment one labeled 'Model data' and the other 'Deviation'. I then subtracted the model data readings from the corresponding given data to get the individual deviations. I then averaged out the numbers in the deviation column. The average deviation turned out to be 3.880975.**
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RESPONSE --> I understand. self critique assessment: 2
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09:14:27 How close is your model to the curve you sketched earlier?
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RESPONSE --> Not too terribly close at all. confidence assessment: 2
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09:14:38 ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.**
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RESPONSE --> I understand. self critique assessment: 2
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09:15:54 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The big thing that surprised me was learning how ineffective FOIL was. Not only does your method give a greater understanding of how the math works, it seems to be alot easier than FOIL to me. self critique assessment: 2
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09:16:08 ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation **
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RESPONSE --> I understand. self critique assessment: 2
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course MTH 163 Enclosed is Week 2 Quiz #1. I randomly chose version 9. All it contanied was the one question. Hope I did this right. The quadratic formula states that if y = ax^2 + bx + c
course MTH 163 Here are the remaining few questions from the q_a_rates assignment on 09-01. Not sure how I missed these to begin with, but I'm sorry for the confusion. }{gЗ֎ދassignment #001
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16:43:28 `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?
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RESPONSE --> If the runner is moving 10 m/s at the 100-meter mark and 9 m/s at the 200-meter mark, this means his rate of speed has changed somewhere during this time. It would take him 10 seconds to cover to length at 10 m/s. It would take him 11.11 seconds to cover the distance at 9 m/s. Therefore, it would take the runner between 10 and 11.11 seconds to cover the distance. confidence assessment: 2
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16:44:52 At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> I see I should've formed an average speed. confidence assessment: 1
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16:45:51 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
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RESPONSE --> Before we weren't looking for an average. We were measuring a change in two points. confidence assessment: 2
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16:46:10 In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
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RESPONSE --> I understand. confidence assessment: 2
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